Find the sum of all the terms in the n-th row of the given series

Find the sum of all the terms in the nth row of the series given below. 

               1  2
            3  4  5  6
         7  8  9 10 11 12
     13 14 15 16 17 18 19 20
    ..........................
   ............................
             (so on)

Examples:

Input : n = 2
Output : 18
terms in 2nd row and their sum
sum = (3 + 4 + 5 + 6) = 18

Input : n = 4
Output : 132

Naive Approach: Using two loops. Outer loop executes for i = 1 to n times. Inner loop executes for j = 1 to 2 * i times. Counter variable k to keep track of the current term in the series. When i = n, the values of k are accumulated to the sum. 
Time Complexity: O(k), where k is the total number of terms from the beginning till the end of the nth row.
Efficient Approach: The sum of all the terms in the nth row can be obtained by the formula: 

 Sum(n) = n * (2 * n2 + 1)

The proof for the formula is given below:
Prerequisite:

1. Sum of n terms of an Arithmetic Progression series with a as the first term and d as the common difference is given as: 
   

   Sum = (n * [2*a + (n-1)*d]) / 2

2. Sum of 1st n natural numbers is given as: 
   

   Sum = (n * (n + 1)) / 2

Proof:

Let the number of terms from the beginning 
till the end of the nth row be p.
Here p = 2 + 4 + 6 + .....n terms
For the given AP series, a = 2, d = 2.
Using the above formula for the sum of
n terms of the AP series, we get,

     p = n * (n + 1)

Similarly, let the number of terms from the 
beginning till the end of the (n-1)th row be q.
Here q = 2 + 4 + 6 + .....n-1 terms
For the given AP series, a = 2, d = 2.
Using the above formula for the sum of
n-1 terms of the AP series, we get,

     q = n * (n - 1)

Now,
Sum of all the terms in the nth row 
           = sum of 1st p natural numbers - 
             sum of 1st q natural numbers
    
           = (p * (p + 1)) / 2 - (q * (q + 1)) / 2

Substituting the values of p and q and then solving
the equation, we will get,

Sum of all the terms in the nth row = n * (2 * n2 + 1)

C++

// C++ implementation to find the sum of all the
// terms in the nth row of the given series
#include <bits/stdc++.h>

using namespace std;

// function to find the required sum
int sumOfTermsInNthRow(int n)
{
    // sum = n * (2 * n^2 + 1)
    int sum = n * (2 * pow(n, 2) + 1);
    return sum;
}

// Driver program to test above
int main()
{
    int n = 4;
    cout << "Sum of all the terms in nth row = "
         << sumOfTermsInNthRow(n);
    return 0;
}

Java

// Java implementation to find the sum of all the
// terms in the nth row of the given series

import static java.lang.Math.pow;

class Test {
    // method to find the required sum
    static int sumOfTermsInNthRow(int n)
    {
        // sum = n * (2 * n^2 + 1)
        int sum = (int)(n * (2 * pow(n, 2) + 1));
        return sum;
    }

    // Driver method
    public static void main(String args[])
    {
        int n = 4;
        System.out.println("Sum of all the terms in nth row = "
                           + sumOfTermsInNthRow(n));
    }
}

Python3

# Python 3 implementation to find
# the sum of all the terms in the
# nth row of the given series
from math import pow

# function to find the required sum
def sumOfTermsInNthRow(n):

    # sum = n * (2 * n^2 + 1)
    sum = n * (2 * pow(n, 2) + 1)
    return sum

# Driver Code
if __name__ == '__main__':
    n = 4
    print("Sum of all the terms in nth row =",
                   int(sumOfTermsInNthRow(n)))

# This code is contributed
# by Surendra_Gangwar

C#

// C# implementation to find the sum of all the
// terms in the nth row of the given series
using System;

class Test {
    // method to find the required sum
    static int sumOfTermsInNthRow(int n)
    {
        // sum = n * (2 * n^2 + 1)
        int sum = (int)(n * (2 * Math.Pow(n, 2) + 1));
        return sum;
    }

    // Driver method
    public static void Main()
    {
        int n = 4;
        Console.Write("Sum of all the terms in nth row = "
                      + sumOfTermsInNthRow(n));
    }
}

// This code is contributed by vt_m.

PHP

<?php
// PHP implementation to find
// the sum of all the terms in
// the nth row of the given series

// function to find the required sum
function sumOfTermsInNthRow($n)
{

    // sum = n * (2 * n^2 + 1)
    $sum = $n * (2 * pow($n, 2) + 1);
    return $sum;
}

    // Driver Code
    $n = 4;
    echo "Sum of all the terms in nth row = ",
                        sumOfTermsInNthRow($n);

// This code is contributed by ajit
?>

Javascript

<script>
// javascript implementation to find the sum of all the
// terms in the nth row of the given series

// function to find the required sum
function sumOfTermsInNthRow( n)
{

    // sum = n * (2 * n^2 + 1)
    let sum = n * (2 * Math.pow(n, 2) + 1);
    return sum;
}

// Driver program to test above

    let n = 4;
    document.write( "Sum of all the terms in nth row = "
         + sumOfTermsInNthRow(n));

// This code is contributed by aashish1995

</script>

Output: 

Sum of all the terms in nth row = 132

Time Complexity: O(1), the code will run in a constant time.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

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